SOLUTION: You measure 22 backpacks' weights, and find they have a mean weight of 47 ounces. Assume the population standard deviation is 10.9 ounces. Based on this, what is the maximal margin

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Question 1199868: You measure 22 backpacks' weights, and find they have a mean weight of 47 ounces. Assume the population standard deviation is 10.9 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean backpack weight.
Give your answer as a decimal, to two places
Answer by ElectricPavlov(122) About Me  (Show Source):
You can put this solution on YOUR website!
**1. Find the Critical Value (z*)**
* For a 95% confidence level, the critical value (z*) is 1.96. This value corresponds to the z-score that captures 95% of the data within two standard deviations of the mean in a standard normal distribution.
**2. Calculate the Standard Error**
* Standard Error (SE) = σ / √n
* where:
* σ = population standard deviation (10.9 ounces)
* n = sample size (22 backpacks)
* SE = 10.9 / √22
* SE ≈ 2.32 ounces
**3. Calculate the Margin of Error**
* Margin of Error = z* * SE
* Margin of Error = 1.96 * 2.32
* Margin of Error ≈ 4.54
**Therefore, the maximal margin of error associated with a 95% confidence interval for the true population mean backpack weight is approximately 4.54 ounces.**